Add leading zero's to string, without (s)printf

Question

I want to add a variable of leading zero\'s to a string. I couldn\'t find anything on Google, without someone mentioning (s)printf, but I want to do this without (s)printf.<

Answer 1

You could do something like:

std::cout << std::setw(5) << std::setfill('0') << 1;

This should print 00001.

Note, however, that the fill-character is "sticky", so when you're done using zero-filling, you'll have to use std::cout << std::setfill(' '); to get the usual behavior again.

Answer 2

memcpy(target,'0',sizeof(target));
target[sizeof(target)-1] = 0;

Then stick whatever string you want zero prefixed at the end of the buffer.

If it is an integer number, remember log_base10(number)+1 (aka ln(number)/ln(10)+1) gives you the length of the number.

Answer 3

// assuming that `original_string` is of type `std:string`:

std::string dest = std::string( number_of_zeros, '0').append( original_string);

Answer 4

The C++ way of doing it is with setw, ios_base::width and setfill

#include <iostream>
#include <iomanip>

using namespace std;

int main()
{
    const int a = 12;
    const int b = 6;

    cout << setw(width) << row * col;
    cout << endl;

    return 0;
}

Answer 5

I can give this one-line solution if you want a field of n_zero zeros:

  std::string new_string = std::string(n_zero - old_string.length(), '0') + old_string;

For example: old_string = "45"; n_zero = 4; new_string = "0045";

Answer 6

This works well for me. You don't need to switch setfill back to ' ', as this a temporary stream.

std::string to_zero_lead(const int value, const unsigned precision)
{
     std::ostringstream oss;
     oss << std::setw(precision) << std::setfill('0') << value;
     return oss.str();
}